When charger 1 is connected to my battery, the charging current (measured by a connecting ammeter in series to the USB cable) is 1.12A and the current used by the battery is 109mA (measured by a connecting ammeter in series to the positive side of the battery)
When charger 2 is connected to my battery, the charging current (measured by a connecting ammeter in series to the USB cable) is 1.332A and the current used by the battery is 110mA (measured by a connecting ammeter in series to the positive side of the battery)

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\$\begingroup\$Welcome - (a) Somewhere in your setup, you must have a battery charger, but there are no details given about that. Please edit the question and add links to its spec and photos of the wiring being used between the power supply, charger and battery. (b) "How can I increase the o/p current of my battery by using charger 2?" Why do you think that using power supply 2 should increase the current to the battery? Although you called them "chargers", since they output 5V (and that can't be used to charge a Li-Ion battery directly) there must be a separate charger somewhere (see point (a)). Thanks.\$\endgroup\$
– SamGibsonJan 29 at 20:33

The thing you are calling a charger is actually just a power supply. The actual charging circuit is part of the battery pack, and it is designed to keep you battery pack from going "boom." That means that (among other things,) it regulates the charge current going into the battery.

The charging circuit will use as much current from the power supply as it was designed to draw. No more. A power supply that provides the proper voltage will not "push" more current through the charger than it is designed for.

Just because the power supply can supply more current, the charger doesn't have to (and won't) draw more current.

A power supply that can supply more current does not mean that your battery must charge faster.

\$\begingroup\$Thank you, I'm using the larger current capacity charger right, Does it will help in improving the actual charge current ?\$\endgroup\$
– user211373Jan 29 at 20:45

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\$\begingroup\$I just said NO in my answer. What part of "you can't" was unclear?\$\endgroup\$
– JREJan 29 at 20:47

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\$\begingroup\$@user211373 - the thing you're calling a 'charger' is not a charger. It's a power supply, and as JRE has explained it does not decide what the actual charging current is.\$\endgroup\$
– brhansJan 29 at 21:25

More Cells in Series Enables Full Current Charges

If you need faster charging at the upper end of the voltage range, you may have to use a different configuration, such as 2S1P, 2S2P, ... The 2S part is 2 cells in series. The nominal voltage in that case would be 3.7 x 2 = 7.4 volts. So, to charge really fast, right up to 4.2 volts, which is the typical max, you actually have to be able to charge up to the 8.4 volts (4.2 x 2), or even 3S at 12.6 volts (4.2 x 3).

Nearly-Charged Batteries Charge Slower in Constant Voltage Mode

Most likely, you have a low charging current because the constant voltage part of the charging curve has already kicked in. This actually means that your cell or battery is already mostly charged. When the battery is its deadest, and you put the battery on the charger (or use your power supply), the protection circuit built in to your cell/battery will limit the charge based upon the actual state of the battery. At first, you will be able to pump some high current into your battery, as it can take that amount of current without actually reaching a voltage that is too high, which is what it absolutely MUST avoid.

How To Test It

To prove what I'm saying, use up the energy in your battery/cell until its energy is gone and it won't give up any more. Use your multimeter to test what the voltage is on the battery. It might be as low as 3.0 volts, which is okay. Then, at that point, charge your battery with both power supplies (or chargers) and notice what current is actually being delivered to the battery/cell at that point -- it should be much more than the approx. 100mA you have already encountered.

Making A Graph Yourself To Better Understand

This should be the constant current phase of the charge. If you keep track of the charge by using a timer, and recording battery voltage, and current being delivered, at 15-minute increments, then you can end up with a nice curve that will show you exactly what is going on, and you should see the constant current phase of the charge, and also see the constant voltage part of the curve, where the current starts to go exponentially down.

Why The Current Ramps Down At End-Of-Charge

The real reason why the charge can not deliver full current during the entire charge is that it would create an overvoltage condition on the battery, because the charge delivered ultimately has to do with the delta-v, the voltage differential.

One Way Danger Plays Out

In other words, if you have unprotected cells or otherwise force a higher current when the battery voltage is already close to full, it would cause a dangerous voltage to appear on the cell, significantly over the 4.2 volt max, causing cell damage by plating of the electrodes, which would start little hairs of metal (dendrites, stalagtites, stalagmites, whiskers, ... different names are used, and different visualizations are useful), but that process which was started eventually causes the whiskers to become long enough to contact the other terminal of the battery, which causes a catastrophic short circuit, and the battery basically explodes just like a flamethrower or a rocket or some type of fireworks.

Using A Fireproof Container For Extra Safety

If you play around with these things, it is useful to put the cell/battery in a fireproof container which can be quickly covered as a precaution -- it would be good not to burn your own house down. I use a steel pot, with top, which I found used for a few bucks. And if you KNOW you're going into the dangerous areas, moving the pot + lid + experiment outside on a cement pad would be prudent.

Wisdom is to do research on this and learn what you are doing, and also to take precautions that are reasonable. Please protect yourself and your loved ones!

After reading how dangerous this can be, it should be easy for you to understand why the protections that are already in place have stopped you from shooting yourself in the foot, so to speak. Otherwise, you would have already created a dangerous situation. I am very thankful for the electrical engineers who do this for a living and work hard to protect us (and their companies from litigation), which is a win-win for everybody involved.

Visit Dave's EEVBlog For Some Great Videos Explaining These Concepts

Now that I've warned you enough, here's one place to go for good information on how to do this and some info. on how to be safe.

I enjoy watching Dave of EEVBlog, and he has some very helpful videos that you should probably watch, as a picture is worth 1,000 words. The following speak directly to your situation and are the most helpful:

It appears that you are using a self-contained device that contains one or more rechargeable cell PLUS a built-in charger for those cells. You specifically mention a 5V supply rail with various current capability.

What you probably fail to understand is that the built-in charger determines how fast the internal cell(s) will charge. That charger will consume as much current as it needs and no more.

The only possible way that you can influence the charge current is if the internal charger has some means of communication with the external power source that is powering the charger. You will often see that with smartphones and such. Charging the phone from a standard 5V supply will often lead to low charging current because the phone doesn't know how much power the source can supply.

But use the power source supplied with the phone and the charge rate is far faster. This is because the phone has communicated with the power source in some fashion and knows that the power source can supply the required current.

In days gone by, this was done by shorting the (D-) & (D+) pins in the USB connector on the power source. This then evolved to putting different voltage levels on each of the (D-) & (D+) pins. The different voltage levels represent different current capabilities.

Unfortunately, the idea of using different voltage levels on the data pins is manufacturer-specific and is different for each manufacturer. For example, only the charger supplied with a specific Apple product will charge at maximum current rate only with that particular product.